Re: [Haskell-cafe] Why is Megaparsec treating these two operators differently?

Awww yiss! Thanks, David! On Sun, May 14, 2017 at 12:13 AM, David Turner < dct25-561bs@mythic-beasts.com> wrote:

The "code in full" link has operators called `#1` and `#2` rather than `|+|` and `|-|`, but I see one of your test cases fail there too. You're using quite an old version of megaparsec, 4.3.0 (stackage lts-5.5) and it looks like something affecting this was fixed in 4.4.0 (e.g. stackage lts-6.9):

$ stack --resolver lts-5.5 ghci --no-load --no-build Configuring GHCi with the following packages: Dwt GHCi, version 7.10.3: http://www.haskell.org/ghc/ :? for help Prelude> :l howto/megaparsec/minimal.hs [1 of 1] Compiling Experim ( howto/megaparsec/minimal.hs, interpreted ) Ok, modules loaded: Experim. *Experim> test Just (Pair (Var "a") (Var "b")) Nothing Just (Pair (Pair (Var "a") (Var "b")) (Pair (Var "c") (Var "d"))) Just (Pair (Pair (Var "a") (Var "b")) (Pair (Var "c") (Var "d"))) *Experim> Leaving GHCi. $ stack --resolver lts-6.9 ghci --no-load --no-build Configuring GHCi with the following packages: Dwt GHCi, version 7.10.3: http://www.haskell.org/ghc/ :? for help Prelude> :l howto/megaparsec/minimal.hs [1 of 1] Compiling Experim ( howto/megaparsec/minimal.hs, interpreted ) Ok, modules loaded: Experim. *Experim> test Just (Pair (Var "a") (Var "b")) Just (Pair (Var "a") (Var "b")) Just (Pair (Pair (Var "a") (Var "b")) (Pair (Var "c") (Var "d"))) Just (Pair (Pair (Var "a") (Var "b")) (Pair (Var "c") (Var "d"))) *Experim> Leaving GHCi.

On 14 May 2017 at 07:53, Jeffrey Brown wrote:

...

The problem's not solved! Check out this weirdness. |+| and |-| are supposed to be identical, mapping to the same function, the Pair constructor.

> mapM_ putStrLn exprs a |+| b a |-| b a |+| b |-| c a |-| b |+| c a |+| b |-| c |+| d > mapM_ putStrLn $ map (show . parseMaybe aExpr) exprs Just (Pair (Var "a") (Var "b")) Nothing Just (Pair (Pair (Var "a") (Var "b")) (Var "c")) Nothing Just (Pair (Pair (Var "a") (Var "b")) (Pair (Var "c") (Var "d")))

Why the two failures?

If I change them from InfixN to InfixR or InfixL, then only the first of those five expressions parses.

Here are two relevant definitions:

data AExpr = Var String | Pair AExpr AExpr deriving (Show)

aOperators :: [[Operator Parser AExpr]] aOperators = [ [ InfixN $ symbol "|+|" *> pure (Pair) ] , [ InfixN $ symbol "|-|" *> pure (Pair) ] -- binds last, I think ]

And here is the code in full[1].

Thanks, Jeff

[1] https://github.com/JeffreyBenjaminBrown/digraphs-with-text/ blob/master/howto/megaparsec/minimal.hs

On Sun, Oct 23, 2016 at 2:50 PM, Brandon Allbery wrote:

...

Aha. I had forgotten some details.

If you want to have an operator that is a prefix of another operator in

...

the table, use the following (or similar) wrapper instead of plain symbol: op n = (lexeme . try) (string n <* notFollowedBy punctuationChar)

http://hackage.haskell.org/package/megaparsec-5.1.1/docs/Tex t-Megaparsec-Expr.html#v:makeExprParser

So you actually need to be a little clever for those two operators to work; it's not as simple as I had recalled it (which would have been correct for a basic manual combinator setup). I am going to guess that something in there is not using `try` and silently consuming the extra "#", but I'd have to study the `makeExprParser` code in Megaparsec to be certain.

On Sun, Oct 23, 2016 at 5:38 PM, Jeffrey Brown wrote:

...

Thanks, Brandon! How did you know that?

I changed them to "#1" and "#2" and now it works[1].

But before making that change, why would "a # b ## c # d" evaluate, even though "a ## b" would not?

[1] https://github.com/JeffreyBenjaminBrown/digraphs-with-text/t ree/master/howto/megaparsec The corrected file is called "experim.hs"; the old one, uncorrected, is called "experim.buggy.hs".

On Sun, Oct 23, 2016 at 2:03 PM, Brandon Allbery wrote:

...

On Sun, Oct 23, 2016 at 4:15 PM, Jeffrey Brown < jeffbrown.the@gmail.com> wrote:

...

[ [ InfixN # symbol "#" *> pure (Pair) ] , [ InfixN # symbol "##" *> pure (Pair) ] ]

Combinator parsers can't rearrange themselves to do longest token matching. So the ## operator will take the first case, match against `symbol "#"` and aOperator will succeed; the the next token match will hit the unconsumed "#" and fail. If you place "##" first then it will match "##" but not "#", which would the match the second rule.

-- brandon s allbery kf8nh sine nomine associates allbery.b@gmail.com ballbery@sinenomine.net unix, openafs, kerberos, infrastructure, xmonad http://sinenomine.net

-- Jeff Brown | Jeffrey Benjamin Brown Website https://msu.edu/~brown202/ | Facebook https://www.facebook.com/mejeff.younotjeff | LinkedIn https://www.linkedin.com/in/jeffreybenjaminbrown(I often miss messages here) | Github https://github.com/jeffreybenjaminbrown

-- brandon s allbery kf8nh sine nomine associates allbery.b@gmail.com ballbery@sinenomine.net unix, openafs, kerberos, infrastructure, xmonad http://sinenomine.net

-- Jeff Brown | Jeffrey Benjamin Brown Website https://msu.edu/~brown202/ | Facebook https://www.facebook.com/mejeff.younotjeff | LinkedIn https://www.linkedin.com/in/jeffreybenjaminbrown(spammy, so I often miss messages here) | Github https://github.com/jeffreybenjaminbrown

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-- Jeff Brown | Jeffrey Benjamin Brown Website https://msu.edu/~brown202/ | Facebook https://www.facebook.com/mejeff.younotjeff | LinkedIn https://www.linkedin.com/in/jeffreybenjaminbrown(spammy, so I often miss messages here) | Github https://github.com/jeffreybenjaminbrown